Let us consider the following boundary value problem:

Аф(г) = 0 r Є V (3.1)

Вф(г) = f (r) r Є dV, (3.2)

where A and В are linear operators. Group theory is not a panacea to the solution of boundary value problems; its application is limited. The main condition that must be met in nuclear engineering problems is that material distributions have symmetry. This is generally true in reactor cores, core cells and cell nodes.

In the following we give a heuristic outline of how the machinery presented above enters into the solution algorithm of a boundary value problem, and what benefits can be expected.

Symmetry is the key. If we have determined that the physical problem has symmetries these symmetries must form a group G. The symmetry operator Og must commute for all g Є G with the linear operators A and В for group theory to be applicable. That is

Og A = AOg and Og В = BOg (3.3)

must hold for all g Є G. If this condition is met, the boundary value problem can be written

as

AOgf(r) = 0 r Є V (3.4)

BOgf(r) = Ogf(r) r Є dV. (3.5)

We can now use the projection operator (2.20) to form a set of boundary value problems

APxf(r) = 0 r Є V (3.6)

BP>(r) = Paf (r) r Є dV. (3.7)

Since the projection operator creates linearly independent components, we have decomposed the boundary value problem into a number (equal to the number of irreducible components) of independent boundary value problems. These are

Atfx (r) = 0 r Є V (3.8)

Вфа (r) = fa (r) r Є dV, (3.9)

whose solution (r) belongs to the a-th irreducible representation. From this complete set of

linearly independent orthogonal functions we reconstruct the solution to the original problem as

nc

Hr) = E (r), (3.10)

a=1

where nc is the number of classes in G.

Why is this better? Recall that we are applying harmonic analysis. The usual approach is to use some series that forms an incomplete set of expansion functions and results a coupled set of equations; one "large" matrix problem. With group theory, we find a relatively small set of complete basis functions that form the solution from symmetry considerations. These are found by solving a set of "small" boundary value problems. It is clear that the effectiveness of group theory is problem dependent. However, experience over the past half century has proven group theory’s effectiveness in both nuclear engineering and other fields.

We present an especially simple example (Allgover et al., 1992) that demonstrates the advantages of symmetry considerations. The example is the solution of a linear system of equations with six unknowns:

1 5 б 2 3 4		•и		9
5 1 4 3 2 б		Х2		14
3 4 1 5 б 2		•3		21
2 б 5 1 4 3		•4		15
б 2 3 4 1 5		•5		14
4 3 2 б 5 1		•б		11

The example has been constructed so that the basis of the reduction is the observation that the matrix is invariant under the following permutations: pi = (1,6)(2,5)(3,4) and p2 = (1,5,3) (2,6,4). As pi and p2 generate a group Об of six element, the matrix commutes with the representation of group Об by matrices of order six. This suggests the application of group theory: decompose the matrix and the vector on the right hand side of the equation into irreducible components, and solve the resulting equations in the irreducible subspaces. The Об group is isomorphic to the symmetry group of the regular triangle discussed in Section 2.2.

The character table of the group Об can be found in tables (Atkins, 1970; Conway, 2003; Landau & Lifshitz, 1980), or, can be looked up in computer programs, or libraries (GAP, 2008).

Using the character table, and projector (2.17), one can carry out the following calculations. The observation that Об is isomorphic to the symmetry group of the equilateral triangle makes the problem easier. (Mackey, 1980) has made the observation: There is an analogy of the group characters and the Fourier transform. This allows the construction of irreducible vectors by the following ad hoc method. Form the following N-tuples (N = |G|):

Є2Ц = (cos(2n/N * (2k — 1) * 1),…, cos(2n/N * (2k — 1) * N),

e2k = (sin(2n/N * (2k) * 1),… ,sin(2n/N * (2k) * N),k = 1,2,… N. (3.12)

These vectors are orthonormal and can serve as an irreducible basis. After normalization, one gets a set of irreducible vectors in the N copies of the fundamental domain. Here one may exploit the isomorphism with the symmetry group of an equilateral triangle with the points positioned as shown in Fig. 1. Applying the above recipe to the points in the triangle, we get the following irreducible basis:

e1 = (1,1,1,1,1,1) e2 = (2, —1, —1,2, —1, —1) e3 = (0,1, —1,0,1, —1) (3.13)

e4 = (2,1, —1, —2, —1,1) e5 = (0,1,1,0, —1, —1)) еб = (1, —1,1, —1,1, —1). (3.14)

We note that the points in the vectors e; do not follow the order shown in Fig. 1. Thus we need to renumber the points, and normalize the vectors. For ease of interpolation we also renumber the vectors given above. It is clear that the vectors formed from cos and sin transform together. Thus they form a two-dimensional representation. We bring forward the one-dimensional representations. The projection to the irreducible basis is through a б x б matrix that contains

Fig. 1. Labeling Positions of Points on an Orbit the orthonormal e’; vectors:

where the prime indicates rearranging in accordance with Fig. 1. Using the rearranging

Ax = b, OAO-1 (Ox) = Ob,

we find[14]

21 0 0 0 0 0

0 -10 0 0 0

0 0 -6 2a 0 0

0 0 — a -10 0 ,

0 0 0 0 -6 2a

0 0 0 0 — a -1

where a = /3. Compare the structure of the above matrix with that given in Section 3, where the similar form is achieved by geometrical similarity. In the present example there is no geometry, just a matrix invariant under a group of transformations.

In order to solve the resulting equations, we need the transformed right hand side of the equation:

Ob = (l4C6,q/|,o,-8,4,

Finally, note that instead of solving one equation with six unknowns, we have four equations, two of them are solved by one division for each, and we have to solve two pairs of equations with two unknowns for each. At the end, we have to transform back from Ox to x.

The Reader may ask: What is the benefit of the reduction? In a problem which is at the verge of solvability, that kind of reduction may become important. [15]

A more favorable situation is when there are geometric transformations leaving the equation and the volume under consideration, invariant. But before immersing into the symmetry hunting, we investigate the diffusion equation.